Excluded Volume Calculation in Van der Waals Equation When accounting for the excluded volume for in the Van der Waals equation, it is assumed that the molecules are hard spheres and are of diameter. If we consider a cube of volume V, then we can say that the side of this cube is of length $V^{1/3}$. Consider the diameter of the molecules to be $\sigma$. Suppose that the number of molecules in this box to be $N$. If we anchor $N-1$ molecules at their positions and look at the excluded volume from the perspective of the $N^{th}$! molecule, we see that the center of this molecule can approach the walls of the cube only upto a distance of $\sigma/2$ and can approach the anchored molecules upto a distance of $\sigma$ from their centers as shown:.
Then the excluded volume for this molecule should be $V_{ex}=(V^{1/3}-\sigma)^{3}-(N-1)(\frac{4}{3}\pi\sigma^{3})$. This follows even if we consider any other molecule and anchor the rest. But, according to wikipedia, we would be overcounting. I don't see how. The correct expression should be $V_{ex}=(V^{1/3}-\sigma)^{3}-(N/2)(\frac{4}{3}\pi\sigma^{3})$. Can anyone please explain?
 A: From Principles of Colloid and Surface Chemistry by Hiemenz and Rajagopalan (if you get an error about viewing the requested page of the book, try refreshing):

The actual excluded volume per atom, $b'$ ($b$, the excluded volume per mole, is equal to $N_A b'$, with $N_A$ the Avogadro's number) is, however, smaller than $\frac{4}{3}\pi\sigma^3$ since the excluded volume of an atom as calculated above may overlap with that of other atoms. Therefore, to obtain an expression for $b$, we need to multiply the above value by $N$ (since there are $N$ atoms in the volume), take half of it since otherwise we will be "double counting" the excluded volumes, and divide by $N$ to get excluded volume per atom, that is
$$b' = \frac{4}{3}\pi\sigma^3 \cdot \frac{N}{2} \cdot \frac{1}{N} = \frac{2}{3}\pi\sigma^3$$

The reason for dividing by 2 rather than some other constant is still somewhat unclear, but the overlap explanation at least shows why multiplying $N$ by the volume of a sphere of radius $\sigma$ would be overcounting.
A: As mentioned in the wikipedia page $4 \times \frac{4 \pi r^3}{3}$ is the excluded volume per particle, so you have to sum over all the particles and divide by the number of particles. While summing up you divide by 2, because a pair of particles only contribute once to the excluded volume. 
